Mixed Boundary Value Problems Episode 16 pot

Equation 6.3.32 and Equation 6.3.33 are best when the integrals can be evaluated exactly while Equation 6.3.34 should be used otherwise... Karapetian, 1994: Elementary exact method for s

440 Mixed Boundary Value Problems

or

u(r, θ, z) = 1

π2

 a

0

 2π

0

R

ξ + arctan

ξ R

z

R3f (ρ, ϑ) dϑ ρ dρ, (6.3.19)

where R2= r2+ ρ2− 2rρ cos(θ − ϑ) + z2,

 1,2 (x) = 12



(x + r)2+ z2∓(x − r)2+ z2, (6.3.20)

the operator L(k)σ(r, θ) is given by

L(k)σ(r, θ) = 1

2π

 2π

0

λ(k, θ − ϑ)σ(r, ϑ) dϑ (6.3.21)

=

∞



n =−∞

k |n| e inθ

1

2π

 2π

0

e −inϑ σ(r, ϑ) dϑ , (6.3.22)

λ(k, ϑ) = 1− k2

1 + k2− 2k cos(ϑ) =

∞



n =−∞

k |n| e inϑ

and

ξ = z



a2− ρ2



2(a) − a2 =



2

2(a) − 2

1(ρ)



2

2(a) − 2

2(ρ)

=



a2− ρ2

a2− 2(a)



a2− ρ2

2(a) − r2

2(a) . (6.3.25)

Equation 6.3.18 is recommended in those cases when the integrals can be evaluated exactly while Equation 6.3.19 is more convenient when the integrals must be computed numerically

To illustrate Fabrikant’s results, consider the case when f (r, θ) = w0, a

constant In this case,

L(ρ)f (ρ, θ) =

∞



n =−∞

ρ |n| e inθ

1

2π

 2π

0

e −inϑ w

0dϑ = w0 (6.3.26)

because all of the terms in the summation vanish except n = 0 Therefore,

d dη

 η

0

L(ρ)f (ρ, θ) ρ

η2− ρ2dρ



= w0, (6.3.27)

and

L

2(η)

rη2

d dη

 η

0

L(ρ)f (ρ, θ) ρ

η2− ρ2dρ



(6.3.28)

=

∞



n =−∞

2(η)

rη2

|n|

e inθ

1

2π

 2π

−inϑ w

0dϑ = w0.

Green’s Function 441 From Equation 6.3.18

u(r, θ, z) = 2w0

π

 a

0

d1(η)



r2− 2

1(η)

= 2w0

π arcsin

2(η)

r a

0

(6.3.29)

=2w0

π arcsin

2(a)

Fabrikant also considered the case when the mixed boundary condition reads

u z (r, θ, 0) = σ(r, θ), 0≤ r < a, 0 ≤ θ < 2π,

u(r, θ, 0) = 0, a < r < ∞, 0 ≤ θ < 2π (6.3.31)

In this case he showed that the solution is

u(r, θ, z) = 4C

 2(a)

2 (0)

 g (x)

0

ρ dρ



g2(x) − ρ2L

ρr

x2

σ(ρ, θ)



dx

√

x2− r2,

(6.3.32)

u(r, θ, z) = 4C

 a

0

 η

0

ρ dρ



η2− ρ2L

ρr

2(η) σ(ρ, θ)



d2(η)



2(η) − r2, (6.3.33)

or

u(r, θ, z) = 2C

π

 a

0

 2π

0

arctan ξ

R

σ(ρ, ϑ)

R dϑ ρ dρ, (6.3.34) where g(x) = x

1 + z2/(r2− x2) and R and ξ have been defined earlier.

The constant coefficient C equals −1/(2π) in classical potential problems and

different values in other applications Equation 6.3.32 and Equation 6.3.33 are best when the integrals can be evaluated exactly while Equation 6.3.34 should be used otherwise

To illustrate Equation 6.3.32 through Equation 6.3.34, consider the case

when σ(r, θ) = σ0, a constant Then,

L

ρr

x2

σ(ρ, θ) =

∞



n =−∞

ρr

x2

|n|

e inθ

1

2π

 2π

0

e −inϑ σ

0dϑ = σ0, (6.3.35)

0

ρ dρ



g2(x) − ρ2L

ρr

x2

σ(ρ, θ) = σ0g(x). (6.3.36)

Therefore, using Equation 6.3.32,

u(r, θ, z) = 4Cσ0

 2(a)

2 (0)

g(x) √ dx

= 4Cσ0

 2(a)

2 (0)

√

x2− r2− z2

442 Mixed Boundary Value Problems

= 2Cσ0

 2(a)

2 (0)

d(x2− r2)

√

x2− r2− z2

− 2Cσ0z2 2(a)

2 (0)

d(x2− r2)

(x2− r2)√

x2− r2− z2 (6.3.39)

= 4Cσ0



x2− r2− z22(a)

2 (0)

− 4Cσ0z arctan

√

x2− r2− z2

z







2(a)

2 (0)

(6.3.40)

= 4Cσ0



a2− 2(a) − 4Cσ0z arctan



a2− 2(a)

z



, (6.3.41)

where 2(0) = r2+ z2and 2(a) + 2(a) = a2+ r2+ z2 Fabrikant has extended

his work to spherical coordinates12and crack problems.13

12 Fabrikant, V I., 1987: Mixed problems of potential theory in spherical coordinates.

Z Angew Math Mech., 67, 507–518.

13 Fabrikant, V I., and E N Karapetian, 1994: Elementary exact method for solving

mixed boundary value problems of potential theory, with applications to half-plane contact

and crack problems Quart J Mech Appl Math., 47, 159–174.

Chapter 7 Conformal Mapping

Conformal mapping is a method from classical mathematical physics for

solv-ing Laplace’s equation It is readily shown that an analytic function w =

ξ + iη = f (z), where z = x + iy, transforms Laplace’s equation in the xy-plane into Laplace’s equation in the ξη-xy-plane The objective here is to choose

a mapping so that the solution is easier to obtain in the new domain This method has been very popular in fields such as electrostatics and hydrodynamics In the case of mixed boundary value problems, this technique has enjoyed limited success because the transformed boundary conditions are very complicated and the corresponding solution to Laplace’s equation is dif-ficult to find In this chapter we illustrate some of the successful transforma-tions

7.1 THE MAPPING z = w + a log(w)

During their study of fringing fields in disc capacitors, Sloggett et al.1 used

this mapping to find the potential in the upper half-plane y > 0 where the potential equals V along the line y = πa when −∞ < x < a ln(a) − a Along

1 Sloggett, G J., N G Barton, and S J Spencer, 1986: Fringing fields in disc

capaci-tors J Phys., Ser A, 19, 2725–2736.

444 Mixed Boundary Value Problems

0

0.5

1

1.5

2

2.5

3

3.5

4

− 0

0.5

1 1.5

0.005

0.025

0.1

0.3

1

1.5

2

3

x

a = 2/ π

Figure 7.1.1: The conformal mapping z = w + a log(w) with a = 2/π If w = ξ + iη, the

dark (solid) lines are lines of constant ξ while the lighter (dashed) lines give η The heavy

dark line corresponds to the line−∞ < ξ < 0 − and η = 0.

y = 0, u(x, 0) = 0 Figure 7.1.1 illustrates this mapping In particular, the following line segments are mapped along the real axis in the w-plane:

−∞ < x < a ln(a) − a y = (aπ)+ −∞ < ξ < −a η = 0

−∞ < x < a ln(a) − a y = (aπ) − −a < ξ < 0 − η = 0

−∞ < x < 1 y = 0 0+< ξ < 1 η = 0

1 < x < ∞ y = 0 1 < ξ < ∞ η = 0

Here the (·)+ and (·) − denote points just above or below (·), respectively.

From Poisson’s integral,

u(ξ, η) = V

π

 ∞

0

η (ξ − s)2+ η2ds =

V

π − V

πarctan

η ξ

(7.1.1)

Therefore, for a given value of ξ and η we can use Equation 7.1.1 to compute

the potential Then the conformal mapping provides the solution for the

corresponding x and y. Figure 7.1.2 illustrates this solution For a given z, Newton’s method was used to solve for w Then the potential follows from

Equation 7.1.1

446 Mixed Boundary Value Problems

0.2

0.4

0.6 0.8

1

1.25 1.5

0.2 0.4

0.6

0.8

1 1.25

1.5

π x/(2b)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Figure 7.2.1: The conformal mapping tanh[πz/(2b)] = sn(w, k) when a/b = 1

2 and k =

tanh[πa/(2b)] The solid, dark line gives values of (w) while the dashed, horizontal lines

denote(w).

can be used to solve Equation 7.2.1 through Equation 7.2.4, where z = x + iy,

w = ξ+iη, k = tanh[πa/(2b)], and sn( ·, ·) is one of the Jacobi elliptic functions Figure 7.2.1 illustrates lines of constant ξ and η within a portion of the original xy-plane It shows that the original semi-infinite strip has been mapped into a rectangular region with 0 < ξ < K and 0 < η < K , where

K and K  are complete elliptic integrals of the first kind for moduli k and

k =√

1− k2.

Applying the conformal mapping Equation 7.2.5, the problem becomes

∂2u

∂ξ2 +

∂2u

∂η2 = 0, 0 < ξ < K, 0 < η < K  , (7.2.6)

subject to the boundary conditions

u(ξ, 0) = 0, u(ξ, K  ) = 1, 0 < ξ < K, (7.2.7)

and

u ξ (0, η) = u ξ (K, η) = 0, 0 < η < K  . (7.2.8)

The solution to Equation 7.2.6 through Equation 7.2.8 is simply u(ξ, η) = η/K  Therefore, lines of constant η/K  give u(x, y) via Equation 7.2.5. Figure

1.1.4illustrates the solution

Conformal Mapping 447

+1

−1

S

S

d

d

z−plane

w−plane

y

η

ξ

x

λ y=

Figure 7.3.1: The conformal mapping used to map the half-plane above the boundary

conditions S m ∪ S d into the half-plane η > 0 in the w-plane.

w2− 1 This mapping is useful in converting an elliptic shaped domain in the xy-plane into a rectangular one in the ξη-plane where w = ξ + iη To illustrate this

conformal mapping, let us solve Laplace’s equation in the half-plane

∂2u

∂x2 +

∂2u

∂y2 = 0, −∞ < x < ∞, η(x, 0) < y < ∞, (7.3.1)

subject to the boundary conditions

lim

y →∞ u(x, y) → 0, −∞ < x < ∞, (7.3.2)

and

∂u

∂n





S d

We begin by solving the problem

∂2u

∂ξ2 +

∂2u

∂η2 = 0, −∞ < ξ < ∞, 0 < η < ∞, (7.3.4)

with the boundary conditions

lim

|ξ|→∞ |u(ξ, η)| < ∞, −∞ < η < ∞, (7.3.5)

448 Mixed Boundary Value Problems

0

0.5

1

1.5

0.7 0.7 0.7 0.7 0.7 7

0.8 0.8 0.8

0.8

0.9 0.9 0.9

0.9

0.99

0 99 0.99 0 99

x

C/V = 0.25

0 0.5 1 1.5

2

0.3 0.3 0.3 0.3 0.3 0.3

0.4 0.4 0.4 0.4 0.4 0.4

0.5 0.5 0.5 0.5 0.5 0.5

0.7 0.7 0.7

0.7

0 99

0 99

99 0 99

x

C/V = 0.50

0

0.5

1

1.5

2

0.001 0.001 0.001 0.001 0.001 001

0.

0.1 0.1 0.1 0.1 0.1

0.25 0.25 0.25 0.25 0.25 0.25

0.5 0.5 0.5

0.5

0.75 0.75 75

0.75

0 9

0 99

0 99 0 99

x

C/V = 0.75

0 0.5 1 1.5 2

1e−20 1e−20 1e−20

e−20 1e−20 1e−20 0.25 0.25

0.25 0.25

0.5 0.5 0.5

0.5

0.75 0.75 0.75

0.75

0 9

0 99

0 99

x

C/V = 1.00

Figure 7.3.2: Plots of the potential u(x, y)/V when λ = 0.5 for various values of C/V

lim

η →∞ u(ξ, η) → 0, −∞ < ξ < ∞, (7.3.6)

u η (ξ, 0) = 0, |ξ| < 1,

The solution to this problem is

u(ξ, η) = V + C i

because

u(ξ, 0) = V + C i

ξ2− 1 = V (7.3.9)

if|ξ| > 1, and

∂u(ξ, 0)

∂η = C 

− ξ

ξ2− 1

if|ξ| < 1 Therefore,

u(x, y) = V + C 

i −λz + √ z2+ λ2− 1

1− λ2

where z = x + iy and C is a free parameter.

Figure 7.3.2 illustrates u(x, y) when λ = 0.5 In the construction of the

conformal mapping and solution, it is important to take the branch cut of

√

w2− 1 so that it lies along the real axis in the complex w-plane.

Conformal Mapping 449

Let us solve3 Laplace’s equation in a domain exterior to an infinitely long

cylinder of radius a

∂2u

∂r2 +

1

r

∂u

∂r +

1

r2

∂2u

∂θ2 = 0, a ≤ r < ∞, 0 < |θ| < π, (7.4.1)

subject to the boundary conditions

lim

r →∞ u(r, θ) → 0, 0≤ |θ| ≤ π, (7.4.2)

u r (a, θ) = 1, 0≤ |θ| < α, u(a, θ) = 0, α < |θ| < π (7.4.3)

We begin by introducing the conformal mapping:

w = ˜ ξ + i˜ η = ia z − a

where z = x + iy Equation 7.4.1 then becomes

∂2u

∂ ˜ ξ2 +

∂2u

∂ ˜ η2 = 0, −∞ < ˜ξ < ∞, 0 < ˜η < ∞, (7.4.5)

with the boundary conditions

lim

˜

η →∞ u( ˜ ξ, ˜ η) → 0, −∞ < ˜ξ < ∞, (7.4.6)

u η˜( ˜ξ, 0) = 2a2/( ˜ ξ2+ a2), |˜ξ| < a tan(α/2), u( ˜ ξ, 0) = 0, |˜ξ| > a tan(α/2) (7.4.7)

We now nondimensionalize ˜ξ and ˜ η as follows:

ξ =

˜

a tan(α/2) =− sin(θ)

tan(α/2)

2ar

r2+ a2+ 2ar cos(θ) , (7.4.8)

and

a tan(α/2) = cot(α/2)

r2− a2

r2+ a2+ 2ar cos(θ) . (7.4.9)

Figure 7.4.1illustrates this conformal mapping Equation 7.4.5 then becomes

∂2u

∂ξ2 +

∂2u

∂η2 = 0, −∞ < ξ < ∞, 0 < η < ∞, (7.4.10)

3 See Iossel’, Yu Ya., 1971: A mixed two-dimensional stationary heat-conduction

prob-lem for a cylinder J Engng Phys., 21, 1145–1147.

Conformal Mapping 451

−2

−1 0 1 2

−2

−1 0 1 2

0.5

1

1.5

x/a y/a

Figure 7.4.2: The solution of Equation 7.4.1 subject to the mixed boundary conditions

Equation 7.4.2 and Equation 7.4.3 when α = π/4.

To solve these dual integral equations, we define A(k) by

A(k) =

 1

0

g(t)J0(kt) dt. (7.4.16)

A quick check shows that Equation 7.4.16 satisfies Equation 7.4.15 identically

On the other hand, integrating Equation 7.4.14 with respect to ξ, we find that

 ∞

0

A(k) sin(kξ) dk = −2a arctan[ξ tan(α/2)], |ξ| < 1 (7.4.17)

Next, we substitute Equation 7.4.16 into Equation 7.4.17, interchange the order of integration, then apply Equation 1.4.13 and obtain

 ξ

0

g(t)



ξ2− t2dt = −2a arctan[ξ tan(α/2)], |ξ| < 1 (7.4.18)

Applying the results from Equation 1.2.13 and Equation 1.2.14,

g(t) = − 4a

π

d dt

 t

0

ζ arctan[ζ tan(α/2)]



t2− ζ2 dζ



(7.4.19)

=−2at tan(α/2)

1 + t2tan2(α/2) . (7.4.20)

Finally, if we substitute Equation 7.4.16 and Equation 7.4.20 into Equation 7.4.13,

452 Mixed Boundary Value Problems

u(ξ, η) = −2a tan(α/2)

 1

0

t



1 + t2tan2(α/2)

 ∞

0

J0(kt)e −kη cos(kξ) dk dt

(7.4.21)

=− √ 2 a tan(α/2)

×

 1

0

t

(

t2+ η2+ ξ2+

(t2+ η2− ξ2)2+ 4η2ξ2

[(t2+ η2− ξ2)2+ 4ξ2η2] [1 + t2tan2(α/2)] dt (7.4.22)

=− √ 2 a ln



tan2

α

2

 

(β2− γ2) tan2α

2



+ 2δβ2+ β2tan2α

2



+ δ

tan2α

2

 

(β2− γ2) tan2α

2



+ 2δβ1+ β1tan2α

2



+ δ



 ,

(7.4.23)

where β1 = 2η2, β2 = 1 + η2− ξ2+

(1 + η2− ξ2)2+ 4ξ2η2, γ = 2ξη, and

δ = 1 + (ξ2− η2) tan2(α/2) InFigure 7.4.2we illustrate the solution when

α = π/4.

Let us solve Laplace’s equation in a domain illustrated inFigure 7.5.1

∂2u

∂x2 +

∂2u

∂y2 = 0,

−b < x < 0, −∞ < y < ∞,

0 < x < ∞, 0 < y < ∞, (7.5.1)

subject to the boundary conditions

lim

y →∞ |u(x, y)| < ∞, −b < x < ∞, (7.5.2)

lim

y →−∞ |u(x, y)| < ∞, −b < x < 0, (7.5.3)





u( −b, y) = T0, −∞ < y < ∞, u(0, y) = T0, −∞ < y < 0,

−u y (x, 0) + hu(x, 0) = 0, 0 < y < ∞.

(7.5.4)

In the previous problem we used conformal mapping to transform a mixed boundary value problem with Dirichlet and/or Neumann boundary conditions into a simple domain on which we still have Dirichlet and/or Neumann condi-tions In the present problem, Strakhov4 illustrates the difficulties that arise

in a mixed boundary value problem where one of the boundary conditions is

a Robin condition Here he suggests a method for solving this problem

4 See Strakhov, I A., 1969: One steady-state heat-conduction problem for a polygonal

region with mixed boundary conditions J Engng Phys., 17, 990–994.

454 Mixed Boundary Value Problems

Figure 7.5.2: The conformal mapping given by Equation 7.5.5 The solid lines are lines

of constant ξ, while the dashed lines are lines of constant η.

where  = 2hb/π.

Motivated by the work of Stokes5 and Chester,6we set

where Ψ(w) is an analytic function in any finite portion of the first quadrant

of the w-plane Therefore, the boundary conditions become

[Ψ(iη)] = T0, 0 < η < ∞, (7.5.12)

and



−i(w2+ 1)dΨ(w)

dw + w

2Ψ(w) 

w =ξ

= 0, 0 < ξ < ∞ (7.5.13)

At this point we introduce the function χ(w) defined by

Ψ(w) = χ(w) − 2iT0

5 Stoker, J J., 1947: Surface waves in water of variable depth Quart Appl Math., 5,

1–54.

6 Chester, C R., 1961: Reduction of a boundary value problem of the third kind to one

of the first kind J Math Phys (Cambridge, MA), 40, 68–71.

Conformal Mapping 455

Why have we introduced χ(w)? Because [−2iT0log(w)/π] satisfies Laplace’s

equation and the boundary conditions Equation 7.5.12 and Equation 7.5.13

as |w| → ∞ in the first quadrant of the w-plane, we anticipate that χ(w) = O(1/ |w|) as |w| → ∞ and 0 ≤ arg(w) ≤ π/2.

Introducing Equation 7.5.14 into Equation 7.5.12 and Equation 7.5.13,

we find that

[χ(iη)] = 0, 0 < η < ∞, (7.5.15)

and



−i(w2+ 1)dχ(w)

dw + w

2χ(w) 

w =ξ

=2T0

π

ξ2+ 1

ξ , 0 < ξ < ∞;

(7.5.16)

or

−i(w2+ 1)dχ(w)

dw + w

2χ(w) = 2T0

π

w2+ 1

w + iαT0, (7.5.17)

where α is a free constant Integrating Equation 7.5.17,

χ(w) = T0 w − i

w + i

/2

e −iw ∞

w

ζ − i

ζ + i

/2

e iζ

− 2i

πζ +

α

ζ2+ 1 dζ.

(7.5.18)

We choose those branches of (w −i) /2and (w + i) /2that approach ξ /2along

the positive real axis as w → ∞ The integration occurs over any path in the first quadrant of the w-plane that does not pass through the points w = 0 and w = i.

We must check and see if Equation 7.5.15 is satisfied Let w = iη, 1 <

η < ∞ Then,

χ(iη) = −iT0 η − 1

η + 1

/2

e η

 ∞

η

τ + 1

τ − 1

/2

e −τ 2

πτ +

α

τ2− 1 dτ

(7.5.19)

and[χ(iη)] = 0 for 1 < η < ∞ Therefore, Equation 7.5.15 is satisfied Consider now w = iη with 0 < η < 1 We rewrite Equation 7.5.18 as

χ(w) = T0 w − i

w + i

/2

e −iw

B + 2i π

 w

0

ζ + i

ζ − i

/2

− e iπ/2

e iζ dζ ζ

− e iπ/2 ∞

w

e iζ dζ

ζ − α

 w

0

ζ + i

ζ − i

/2

e iζ dζ

ζ2+ 1



, (7.5.20)

where we have introduced

B = − 2i

π

 ∞

0

ζ + i

ζ − i

/2

− e iπ/2



e iζ dζ

ζ + α

 ∞

0

ζ + i

ζ − i

/2

e iζ dζ

ζ2+ 1.

(7.5.21)

mixed boundary value problems of potential theory, with applications to half-plane contact

and crack problems Quart J Mech Appl Math.,... electrostatics and hydrodynamics In the case of mixed boundary value problems, this technique has enjoyed limited success because the transformed boundary conditions are very complicated and the... Then the potential follows from

Equation 7.1.1